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Let $S = {1, 2, 3, ....... n}$ and $f_{n}$ be the number of those subsets of S which do not contain consecutive elements of S, then

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$f_{n} = 2 f_{n - 1}$

$f_{4} = 8$

$f_{n} = f_{n - 1} + f_{n - 2}$

$f_{n} = \frac{n (n - 1) (n - 2)}{6}$

answer is D.

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Detailed Solution

Let $n = 4$ , then the sunsets of ${1, 2, 3, 4}$ which do not contain consecutive elements of this set and $ϕ$ , ${1}, {2}, {3}, {4}, {1, 3}, {1, 4}, {2, 4}$

$f_{4} = 8 s i m i l a r l y, f = 3, f_{3} = 5$

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